What is ((y+2)^2}{25}-\frac{(x-3)^2)/4 =1 ?

The solution to ((y+2)^2}{25}-\frac{(x-3)^2)/4 =1 is Hyperbola with (h,k)=(3,-2),a=5,b=2

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((y+2)^2}{25}-\frac{(x-3)^2)/4 =1

\frac{( y + 2 ) ^{2}}{25} - \frac{( x - 3 ) ^{2}}{4} = 1

(h, k) = (3, - 2), a = 5, b = 2

Solution steps

\frac{( y + 2 ) ^{2}}{25} - \frac{( x - 3 ) ^{2}}{4} = 1

Rewrite

\frac{( y + 2 ) ^{2}}{25} - \frac{( x - 3 ) ^{2}}{4} = 1

in the form of a standard hyperbola equation

\frac{( y - ( - 2 ) ) ^{2}}{5 ^{2}} - \frac{( x - 3 ) ^{2}}{2 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (3, - 2), a = 5, b = 2

x^{2} - 2 x - 4 y^{2} - 40 y = 103

16 X^{2} - 25 Y^{2} + 96 X + 544 = 0

7 x^{2} - 12 y^{2} - 14 x + 24 y - 28 = 0

(x - 2)^{2} - 4 (y + 2)^{2} = 4

2 x^{2} - y^{2} + 4 x + 4 y - 4 = 0

What is ((y+2)^2}{25}-\frac{(x-3)^2)/4 =1 ?
The solution to ((y+2)^2}{25}-\frac{(x-3)^2)/4 =1 is Hyperbola with (h,k)=(3,-2),a=5,b=2